Please help with Calculus

Question

A pitcher throws a baseball straight into the air with a velocity of 72 feet/sec. If acceleration due to gravity is -32 ft/sec², how many seconds after it leaves the pitcher's hand will it take the ball to reach its highest point? Assume the position at time t = 0 is 0 feet

Options:

1. 2.25
2. 2.5
3. 4.25
4. 4.5

Answer 1

Position equation is y=-16t^2+72t+0 since initial,height is given as 0. this will,be a downward facing parabola and peak height will be at the vertex so all you need to do is find the x value of the vertex which is at -b/2a = -72/-32= 2.25 seconds

Answer 2

The initial velocity, V_i = 72 ft/sec. At the highest point, the ball will temporarily stop before coming back down, and therefore it will have a velocity of 0 at its peak. We will call this the final velocity, V_f.

We will use two formulas to solve this. first we will find out how high the ball travelled.

we use the formula V_f²=V_i²+2*a*(X_f - X_i). where a is acceleration, which equals -32 ft/sec², and X_f is the final heights, and X_i is the initial height (which equals 0 feet as stated in the last sentence of the problem)

plugging in, 0²= 72²+2(-32)(X_f - 0)

simplifying, we get 0 = 5184-64X_f

Rearranging, we get X_f = 5184/64 = 81 feet.

Now we know that its tallest point point is 81 feet.

to find the time, t, in seconds it takes to get there, we use X_f -X_i = 0.5(V_f+V_i)t

plugging in we get, 81-0 = 0.5(0+72)t

simplifying we get 81 = 0.5(72)t = 36t

t = 81/36 = 2.25 seconds